Is Newton's third law of motion consistent with GTR?

In summary,Newtons third law still applies to the forces between apple and branch in GR. But it doesn't apply to gravity acting on the apple, because in GR that is modeled as an inertial force. In GR, inertial forces are not subject to Newtons's 3rd Law, even in classical mechanics.
  • #1
Ahan Sha
3
1
If an apple hanging in the tree has only reaction upwards, then what will happen to a Newtons third law? how is it that there is no "force " downwards, but have spacetime curvature which "mimics" a force. why can't spacetime curvature be itselt a force?
 

Attachments

  • Spacetime_curvature.jpg
    Spacetime_curvature.jpg
    61.3 KB · Views: 451
Physics news on Phys.org
  • #2
First of all, you seem to have a common misunderstanding of Newton's third law. It does not say that the force from the branch on the apple is the reaction force to gravity. The third law partner of the force from the branch on the apple is the force from the apple on the branch. Both of those forces are still there in GR.
 
Last edited:
  • Like
Likes Ahan Sha
  • #3
Ahan Sha said:
If an apple hanging in the tree has only reaction upwards, then what will happen to a Newtons third law? how is it that there is no "force " downwards
As Orodruin wrote, Newtons 3rd still applies to the forces between apple and branch in GR. But it doesn't apply to gravity acting on the apple, because in GR that is modeled as an inertial force. Inertial forces are not subject to Newtons's 3rd Law, even in classical mechanics.

An alternative to using inertial forces is using deformed coordinates:

 
  • Like
Likes Ahan Sha
  • #4
Orodruin said:
First of all, you seem to have a common misunderstanding of Newton's third law. It does not sa that the force from the branch on the apple is the reaction force to gravity. The third law partner of the force from the branch on the apple is the force from the apple on the branch. Both of those forces are still there in GR.[/QUOT
Orodruin said:
First of all, you seem to have a common misunderstanding of Newton's third law. It does not sa that the force from the branch on the apple is the reaction force to gravity. The third law partner of the force from the branch on the apple is the force from the apple on the branch. Both of those forces are still there in GR.

Thanks for your reply.
just
Orodruin said:
First of all, you seem to have a common misunderstanding of Newton's third law. It does not sa that the force from the branch on the apple is the reaction force to gravity. The third law partner of the force from the branch on the apple is the force from the apple on the branch. Both of those forces are still there in GR.
THANKS FOR THE REPLY.
but... my doubt is still unclear. just forget about the branch. Only think about our Apple hanging on that. I thought, Apple is in equilibrium because of gravitational force downwards and reaction force upwards( according to Newtonian Gravity). but you said "It does not say that the force from the branch on the apple is the reaction force to gravity" if this is true, then how is Apple in equilibrium according Newtonian gravity) pls reply
 
  • #5
In Newnonian mechanics, the force on the apple from the branch is equal to the gravitational force on the apple. However, this is just the equilibrium condition. It does not make the forces a third-law pair. The third law partner of the gravitational force on the apple from the Earth is the gravitational force of the Earth on the apple.
 
  • Like
Likes Ahan Sha
  • #6
A.T. said:
Inertial forces are not subject to Newtons's 3rd Law, even in classical mechanics.

I was under the impression that gravity is just as subject to Newton's 3rd Law as any other force is.
 
  • #7
Drakkith said:
I was under the impression that gravity is just as subject to Newton's 3rd Law as any other force is.
He is talking about inertial forces. Gravity is not an inertial force in Newtonian mechanics.
 
  • #8
What about a 2 body system, two objects orbiting about a common center of mass? With Newtonian mechanics, each body exerts a gravitational force on the other, in what could be considered a Newton third law pair. How does GR describe this situation?
 
  • #9
Ahan Sha said:
how is it that there is no "force " downwards, but have spacetime curvature which "mimics" a force. why can't spacetime curvature be itselt a force?

The answer to that is kind of complicated. It has to do with how GR describes gravity as the result of spacetime curvature and how objects react to this curvature. Try this link: http://curious.astro.cornell.edu/ph...force-how-does-it-accelerate-objects-advanced
 
  • #10
Orodruin said:
He is talking about inertial forces. Gravity is not an inertial force in Newtonian mechanics.

Roger.
 
  • #11
rcgldr said:
What about a 2 body system, two objects orbiting about a common center of mass? With Newtonian mechanics, each body exerts a gravitational force on the other, in what could be considered a Newton third law pair. How does GR describe this situation?
Each body is undergoing geodesic motion in space-time. The presence of energy, momentum, and stress affects the shape of space-time and therefore the motion of the bodies.
 
  • #12
rcgldr said:
What about a 2 body system, two objects orbiting about a common center of mass? With Newtonian mechanics, each body exerts a gravitational force on the other, in what could be considered a Newton third law pair. How does GR describe this situation?

Orodruin said:
Each body is undergoing geodesic motion in space-time. The presence of energy, momentum, and stress affects the shape of space-time and therefore the motion of the bodies.
I forgot to ask how GR affects rotating frames of reference. Assume a frame of reference that rotates at the same rate as a 2 body system with a circular orbit (so that the rate of rotation is constant). What is the effect on fictitious centrifugal and coriolis forces?
 
  • #13
rcgldr said:
I forgot to ask how GR affects rotating frames of reference. Assume a frame of reference that rotates at the same rate as a 2 body system with a circular orbit (so that the rate of rotation is constant). What is the effect on fictitious centrifugal and coriolis forces?
This is just putting a different set of coordinates on the same space-time.
 
  • #15
Ahan Sha said:
Only think about our Apple hanging on that. I thought, Apple is in equilibrium because of gravitational force downwards and reaction force upwards( according to Newtonian Gravity).
Whether the Apple is in equilibrium or not is a frame-dependent concept.

In the frame of the Apple an accelerometer at rest measures an acceleration of 1 g upwards, so this frame is a non inertial frame with an inertial force of mg downwards. That inertial force downwards balances the upwards contact force from the branch and the Apple is at equilibrium.

In a local free-fall frame an accelerometer reads 0, so that frame is inertial and there are no inertial forces. The only force on the Apple is the contact force from the branch which accelerates the Apple upwards. So the Apple is not in equilibrium in that frame.
 
  • #16
Orodruin said:
He is talking about inertial forces. Gravity is not an inertial force in Newtonian mechanics.
That depends on your notion of "inertial force"; I'd say it is :P
 
  • #17
Ahan Sha said:
...I thought, Apple is in equilibrium because of gravitational force downwards and reaction force upwards( according to Newtonian Gravity). but you said "It does not say that the force from the branch on the apple is the reaction force to gravity" if this is true, then how is Apple in equilibrium according Newtonian gravity) pls reply

From the Newtonian point of view, the Earth is exerting a gravitational force, F1, downward on the apple and, from the 3rd law, the apple exerts an equal and opposite force, F2, upward on the earth. Also, the apple is exerting a force, F3, downward on the branch and, from the 3rd law, the branch is exerting an equal and opposite force, F4, upward on the apple. Now let's focus on the apple. Because it is in equilibrium, |F4| = |F1| i.e. they are numerically equal other than a sign. But they are not the forces to which the 3rd law applies.
 
  • #18
Orodruin said:
Gravity is not an inertial force in Newtonian mechanics.
haushofer said:
That depends on your notion of "inertial force"; I'd say it is :P
Which notion of "inertial force" includes Newtonian Gravity?
 
  • #19
Suppose one is in EInstein's elevator. Which is one of the usual pedagogical approaches to explaining how gravity works in GR. Would one say that "Newton's third law" still works in the elevator?

Personally, I would not, though I'm not sure how well defined the question is. What I would say is that none of Newton's laws apply directly in an accelerated frame of reference such as the elevator. It's not that the laws are wrong, it's just that they need to be applied in the proper manner, and the proper manner is to apply them in an inertial frame of reference. I would also say that inertial forces are not generally regarded as being real forces, so that if one is standing in an elevator, there is only one real force, that is the force on one's feet pushing you up to make you accelerate along with the elevator.

Now, I recall plenty of lectures (from as far back as high school physics) that "inertial forces are not real forces" but do not have any references handy on the issue. My understanding and recollection though, is that the reasoning is that real forces must transform as tensors, and inertial forces do not. It turns out that inertial forces transform as Christoffel symbols. Of course this isn't the understanding I had in high school, that understanding has developed over time. I believe that the difference between Christoffel symbols and forces is more apparent in generalized coordinates, but some differences are still there even in Cartesian coordinates.. For instance, a tensor that is zero in one coordinate system is zero in all coordinate systems - but inertial "forces" are present only in accelerated coordinate systems and not present in non-accelerated ones.

Having noted the distinction between inertial forces (represented by rank 1 tensors) and non-inertial forces (represented by non-tensorial Christoffel symbols), one logically concludes that gravity is a force in Newtonian mechanics, and not a force (but a non-inertial force) in GR.

The issue I see for B and I level students is making clear the distinction between inertial forces and non-inertial forces. I believe I'm recalling all this correctly, but it's been so long it's hard to be sure, and I am not sure where to check to make sure I haven't forgotten something important or added something that's a bit non-standard. There's also the issue that my explanation above used tensors, which may not be suitable for the target audience :(.

Perhaps the clearest route (and it's not as clear as I'd like) is to follow in Einstein's footstep, and go from the acceleating elevator in Newtonian mechanics to the accelerating elevator in special relativity. In Newtonian mechanics, we pay lip service to the difference between inertial forces and real forces, but the matter didn't seem terribly urgent at the time it was being taught. When we move on to special relativity, though, the differences become much more apparent, and the reasons for all those earlier, not terribly-well understood at the time cautions, becomes more plain in hindsight.

Consider the issue of time dilation. It's well known (though I'm not aware of any really basic non-tensor treatment) that if you have a pair of clocks in an accelerating elevator, they do not stay synchronized. Why does this happen? If inertial forces were actually forces, it wouldn't happen, as forces do not cause time dilation. But the coordinate transformation to an accelerated frame of reference (formally involving the Christoffel symbols) has properties that simply cannot be modeled by a force. So we're back to saying that the fundamental issue is to avoid conflating inertial forces and real forces. They may appear similar in Newtonian mechanics, but when one moves on to even special relativity, the difference between the two concepts becomes more apparent, and one will become confused if one does not make the proper distinctions between inertial forces and real forces.

And the corollary to all this is that while gravity is a real force in Newtonian physics, it's an inertial force in General Relativity.
 
  • Like
Likes Ibix
  • #20
pervect said:
none of Newton's laws apply directly in an accelerated frame of reference such as the elevator.

They do if you are willing to include "inertial forces" in your analysis. See below.

pervect said:
if one is standing in an elevator, there is only one real force, that is the force on one's feet pushing you up to make you accelerate along with the elevator.

No, there is also another real force, the force of your feet pushing back down on the floor of the elevator. As others have pointed out, this force is the one that is paired with the force of the elevator on your field by Newton's Third Law. So Newton's Third Law applies just fine.

Newton's Second Law also applies just fine in the non-inertial elevator frame, provided, as I said above, that you are willing to include an "inertial force" in this frame, which points downward (i.e., towards the elevator floor) and has a magnitude equal to the acceleration of the elevator times the mass of whatever object is being accelerated by it. So, for example, if you are standing on the floor of the elevator and let go a rock, the rock will be acted on by this inertial force, which causes it to accelerate downward. And the magnitude of this force obeys Newton's Second Law, ##F = ma##.

Furthermore, if we want to explain why you, standing on the floor of the elevator, don't accelerate upward as a result of the force the floor exerts on you, we have to appeal to this same inertial force, which acts downward and has a magnitude exactly equal to that of the upward force of the floor on you. So the net force on you is zero, and you remain at rest in this non-inertial frame.

Of course some will object that all of the above is true only because we defined the "inertial force" in order to make it true. This is correct; but it doesn't make the above analysis invalid. It just makes it conceptually limited; we obviously can't use the above analysis to argue that inertial forces "must be real" (or words to that effect), because that would be arguing in a circle.

pervect said:
Having noted the distinction between inertial forces (represented by rank 1 tensors) and non-inertial forces (represented by non-tensorial Christoffel symbols)

I think you mean this the other way around, correct?
 
  • #21
PeterDonis said:
No, there is also another real force, the force of your feet pushing back down on the floor of the elevator. As others have pointed out, this force is the one that is paired with the force of the elevator on your field by Newton's Third Law. So Newton's Third Law applies just fine.

If you drop a rock inside an upward-accelerating elevator, then the rock will "fall" to the floor as if it were acted on by a force. However, there is no Third Law force counter to this fictitious force. The rock experiences a downward force, but there is no equal and opposite force on anything. So I would say that the Third Law is not valid in an accelerating frame if you consider fictitious forces.

Of course, there is a way to do Newtonian gravity that makes it seem almost like General Relativity, in that it views the force of gravity to be a connection coefficient, rather than a real force. That's the Newton-Cartan theory, which I think is equivalent to the usual Newtonian physics.
 
  • Like
Likes Ahan Sha
  • #22
stevendaryl said:
I would say that the Third Law is not valid in an accelerating frame if you consider fictitious forces.

One can actually define "fictitious forces" this way in Newtonian physics--they are forces that don't obey Newton's Third Law. But it's still true that there are forces in accelerating frames that do obey the Third Law, and they are precisely the forces that don't go away when you switch to an inertial frame. (Gravity is a special case--see below.)

stevendaryl said:
That's the Newton-Cartan theory, which I think is equivalent to the usual Newtonian physics.

Except that in this formulation, gravity is a fictitious force by the above definition, whereas in standard Newtonian physics, it isn't.
 
  • #23
A.T. said:
Which notion of "inertial force" includes Newtonian Gravity?
A force which can always be made to dissappear when you go to an appropriate frame of reference. In this case an accelerating one. I don't see why in Newtonian gravity we can't call all the frames of reference with arbitrary time-dependent accelerations "inertial frames"; they are just freely falling. This extends the Galilei group to the Galilei-accelerated ones, just as in GR the Poincaré transformations are extended to general coordinate transformations.

If one argues that inertial forces are due to the non-tensorial character (under Galilei transformations) of the Newtonian equations of motion and don't have a physical origin as described by Newton's second law (a force introduces an acceleration, not the other way around), then I guess indeed one would call gravity a "real force". But it's a matter of interpretation.
 
  • #24
haushofer said:
A force which can always be made to dissappear when you go to an appropriate frame of reference.
Newtonian Gravity is defined as an interaction between two bodies. Can you make the equal but opposite forces on both bodies disappear in some frame of reference frame?
 
  • #25
haushofer said:
A force which can always be made to dissappear when you go to an appropriate frame of reference.

When you look at gravity as a local theory, yes, you can do this. If you want to make it a global theory a la classical Newtonian gravity, this is no longer possible. You can make the gravitational field zero at a given point, but not globally, and globally there is a 3rd law pair to each gravitational force.
 
  • #26
PeterDonis said:
One can actually define "fictitious forces" this way in Newtonian physics--they are forces that don't obey Newton's Third Law. But it's still true that there are forces in accelerating frames that do obey the Third Law, and they are precisely the forces that don't go away when you switch to an inertial frame. (Gravity is a special case--see below.)

Okay, with the exception for fictitious forces, I agree.

Except that in this formulation, gravity is a fictitious force by the above definition, whereas in standard Newtonian physics, it isn't.

But they're equivalent, empirically. Maybe it's analogous to General Relativity vs. spin-two field theory. In the second theory, gravity is not curvature, but the two theories make the same predictions (at least locally).
 
  • #27
PeterDonis said:
Except that in this formulation, gravity is a fictitious force by the above definition, whereas in standard Newtonian physics, it isn't.
As far as I know Newton Cartan makes all of the same experimental predictions as standard Newtonian physics. So I would consider it to be a reformulation of Newtonian physics, not a new theory (despite the treatment of gravity as a fictitious force).
 
  • #28
Orodruin said:
When you look at gravity as a local theory, yes, you can do this. If you want to make it a global theory a la classical Newtonian gravity, this is no longer possible. You can make the gravitational field zero at a given point, but not globally, and globally there is a 3rd law pair to each gravitational force.
Yes, you're right; I forget the important word "local".
 
  • #29
A.T. said:
Newtonian Gravity is defined as an interaction between two bodies. Can you make the equal but opposite forces on both bodies disappear in some frame of reference frame?
I'd say that an observer measures a resultant force due to these two bodys, and can use an acceleration (locally) to consider herself in free fall. That's as far as I know the idea of the equivalence principle.

So the answer to your question is "no". The EP is a local statement.
 
  • #30
what is GTR?
 
  • #31
eltodesukane said:
what is GTR?
General Theory of Relativity.
 
  • #32
Orodruin said:
When you look at gravity as a local theory, yes, you can do this. If you want to make it a global theory a la classical Newtonian gravity, this is no longer possible. You can make the gravitational field zero at a given point, but not globally, and globally there is a 3rd law pair to each gravitational force.

I think this answer is mixing up two different ideas. There is an effective gravitational field that combines gravity with the inertial forces due to choosing noninertial coordinates. This effective field does not have a 3rd law counterpart. If you separate the "real" gravitational field from the inertial forces, then the real field has a 3rd law counterpart.

So my complaint is that you're using "gravitational field" to mean two different things: the effective field (when you say "You can make the gravitational field zero") and the real field at another point (when you say that "globally there is a 3rd law pair to each gravitational force"). I'm perhaps just being nit-picky...
 
  • Like
Likes weirdoguy
  • #33
stevendaryl said:
they're equivalent, empirically.

Yes, agreed. That's one of the reasons for suspecting, even in the absence of any knowledge of GR, that something is fishy about standard Newtonian gravity insisting that gravity is a "real" force.
 
  • Like
Likes haushofer
  • #34
Could one say that causality is an important issue in defining "inertial forces"? E.g., if I'm on a rotating disc, I will measure Coriolis- and centrifugal forces. But I can't identify the force which is acting on the object in order to make it deflect. The only thing I can identify is that there is an engine underneath the rotating disc, making it rotate. That's how I view "inertial forces": usually, you identify the force F on an object, resulting in an acceleration a via F=ma. With inertial forces, you identify the acceleration a (e.g. the deflection of an object), which "results" in a force F (the Coriolis-force). Without a direct cause for this force on the object, Newton's third law won't hold. So, in a way, you reverse the causality behind the equation F=ma.
 
  • #35
haushofer said:
That's how I view "inertial forces": usually, you identify the force F on an object, resulting in an acceleration a via F=ma. With inertial forces, you identify the acceleration a (e.g. the deflection of an object), which "results" in a force F (the Coriolis-force). Without a direct cause for this force on the object, Newton's third law won't hold. So, in a way, you reverse the causality behind the equation F=ma.

Reversing the causality does seem to be one of the major sources of confusion. That needs to be considered when switching from an inertial frame to an accelerated one. Let’s say Object A is being accelerated by surface B in an inertial frame. Call the force that surface B exerts on object A forceBA, and the “reactive” force that object A exerts on Surface B forceAB. If these two 3rd law forces are to be considered “real” forces, then they are also real when switching from an inertial frame to an accelerated one. It seems to me that arguments against this are purely semantic in nature.

In the accelerated frame, the force that object A exerts on surface B is due to its inertia, but instead of calling it an inertial force, let us call it forceAB. Surface B exerts a “reactive” force back on object A, forceBA. These two forces are Newton’s 3rd law pair in the accelerated frame, and match identically with their counterparts in the inertial frame.
 
Last edited:

Similar threads

  • Special and General Relativity
Replies
27
Views
786
  • Special and General Relativity
Replies
14
Views
1K
Replies
17
Views
785
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
165
  • Classical Physics
2
Replies
40
Views
2K
Replies
23
Views
3K
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
776
  • Special and General Relativity
2
Replies
38
Views
2K
Back
Top